"""
Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph. Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.



Example:



Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.


Note:

The number of nodes will be between 1 and 100.
The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
You must return the copy of the given node as a reference to the cloned graph.
"""

"""
# Definition for a Node.
class Node:
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors
"""


class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':

        def bfs(node, visited):
            if node in visited:
                return visited[node]
            if node is None:
                return None
            visited[node] = Node(node.val, [])
            for neighbor in node.neighbors:
                visited[node].neighbors.append(bfs(neighbor, visited))
            return visited[node]

        return bfs(node, {})
    
